TITLE: Is it alright to substitute $0$ for $1/n$ in this limit problem?
QUESTION [7 upvotes]: Evaluate: $\lim\limits_{n \to \infty}\displaystyle\sum_{r=1}^{n} \dfrac{1}{2n + 2r-1}$

To solve this I used the following approach: 
\begin{align} S &= \lim\limits_{n \to \infty}\sum_{r=1}^{n} \dfrac{\frac 1n}{2 + 2\frac rn-\frac1n} \\
&= \lim\limits_{n \to \infty}\displaystyle\sum_{r=1}^{n} \dfrac{\frac 1n}{2 + 2\frac rn-\color{red}{0}} = \dfrac 12\int _0^1 \dfrac{1}{1+x}\,\mathrm{d}x \\ &= \ln(\sqrt 2)\end{align}
Though the answer is correct, I am unsure about my second step in which I have replaced $1/n$ by $0$. Is that allowed? I have tried it in some other problems too and it works.

REPLY [8 votes]: No, this is not allowed, because for all you know that $\frac 1 n$ could have some subtle interaction with the rest of the formula which cancels out its going to zero. For example in the limit
$$\lim_{n\to \infty}n\frac1n$$
we can't just replace $\frac1n$ with $0$ (yielding a limit of $0$) because the presence of the factor $n$ cancels that out. How do you know there isn't something you didn't spot in your formula which produces similar behavior? Maybe the $\frac rn$ or the $\frac 1n$ in the numerator has a similar "cancelling" effect on the $\frac 1n$ which you're trying to replace with $0$.